Determine if function is convex

Question

I want to determine if the function $$f(x,y)=e^x+e^y+x^2-2xy+4y^2+2x+3y-8$$ is convex on $\mathbb{R}^2$. Any ideas on how to prove convexity for two variable functions?

Answer 1

You can show that the hessian is positive semi-definite.

Answer 2

Firstly, a convex function plus/minus a linear function is convex. So you just need to check if $e^x+e^y+x^2-2xy+4y^2$ is convex. It is because it equals $e^x+e^y+(x-y)^2+3y^2$ is a summation of 4 convex functions.

Answer 3

If $u$ and $v$ are convex, so is $u + \lambda v$ for any positive $\lambda.$ Linear functions, constants (negative constants included) and exponential are all convex, so suffices to show $x^2 -2xy+4y^2 = (x-y)^2+3y^2$ is convex, which is obvious (sum of two convex). Is this enough or want a more detailed explanations?

Answer 4

$$f(x,y)=e^x+e^y+x^2-2xy+4y^2+2x+3y-8$$

$$\nabla f=\begin{bmatrix} e^x+2x-2y+2 \\ e^y-2x+8y+3\end{bmatrix} $$

$$\nabla^2 f=\begin{bmatrix} e^x+2 & -2 \\ -2 & e^y+8\end{bmatrix} $$

Since $$e^x+2>0$$

and $$(e^x+2)(e^y+8)-2^2=8e^x+2e^y+16-4=8e^x+2e^y+12>0$$

The function is convex.

Another way to see this is to observe that since

$$\nabla^2 f=\begin{bmatrix} e^x+2 & -2 \\ -2 & e^y+8\end{bmatrix} =\begin{bmatrix} e^x & 0 \\ 0 & e^y\end{bmatrix}+ \begin{bmatrix} 2 & -2 \\ -2 & 8\end{bmatrix}.$$

Observe that $\begin{bmatrix} 2 & -2 \\ -2 & 8\end{bmatrix}$ is positive semidefinite from Gershgorin circle theorem, the eigenvalues lie between $0$ and $10$.

Also, $\begin{bmatrix} e^x & 0 \\ 0 & e^y\end{bmatrix}$ is positive definite.

Hence $\nabla^2f$ is positive definite.